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User talk:Acer4666
Welcome Hi, welcome to Mathematics! Thanks for your edit to the Chain rule page. Please leave a message on my talk page if I can help with anything! -- SpikeToronto (Talk) 13:01, May 21, 2012 I would also like to welcome you, and thank you for the work you've already done. Fallacies like dividing by 0 and the assumption that the square root of x squared is positive should be left in the dust, and I applaud you for doing so. Of all your work on wikis, which makes you the most proud? — Jeff G. ツ 03:52, June 5, 2012 (UTC) About the chain rule The first proof Gottfried Leibniz gave for the chain rule is the precise one I gave, and it is in fact perfectly valid. It is because the limiting operation respects multiplicativity. You just make the proof longer by taking in the complete expression for the derivative. The proof was not fallacious and was the perfect way to obtain the chain rule in the first place. This holds because the two functions are meromorphic in the entire complex plane. The actual analogy of solving differential equations comes from treating "du" as an infinitesimal ''number rather than ''zero. Because, as far as it is small, it is still in existence. While you can't take out just "d" s (because they simply are signs representing infinitesimal change), you can still take out "du" as a whole. And as I said, establishing this fact again with a longer proof is nothing helpful. Your proof, as I said, does not make use of the limit's respect to multiplicativity when both limits exist, which we assume them to.This is how you cancel out differentials "rigorously". Read my new edit on the chain rule page. Starfall.V 15:27, May 31, 2012 (UTC) :No, it's not a proof. Leibniz wrote that but it is by no means a proof - I'm sorry that you find rigour "nothing helpful", but you cannot call it a proof. You haven't defined anything in your proof, you haven't even said what you are cancelling (you just said "we can cancel out 'dy','dx' and other terms" - no specifics given) so there is no way it could ever pass as a proof. Proofs work from definitions and make clear logical steps to the answer. See [here, "it appears that we can just cancel out the du's!!!! (Of course, this really makes no literal sense, since du itself has no explicit numerical significance.", or here, "I gather that you do know that method isn't really a proof.",--Acer4666 (talk) 17:53, May 24, 2012 (UTC) ::I will be happy to inspect any proper proof you give, using definitions and rigour. However your first one was just an arm-waving paragraph with no specifics and what is currently on the page is a load of parsing errors. ::In regards to my proof not making use of the limit's respect to multiplicativity, please read again where I have linked to the Algebra of Limits, that is where I use it.--Acer4666 (talk) 11:47, June 2, 2012 (UTC) Ok, I managed to fix your parsing errors. You were writing "\0" instead of just "0". I'll post the proof here for discussion: "This makes use of the limiting operation's respect to multiplicativity. \frac{\mathrm{d}y}{\mathrm{d}u}\cdot\frac{\mathrm{d}u}{\mathrm{d}x}=\lim_{\delta u \to 0}\frac{y(u+\delta u)-y(u)}{\delta u} \cdot \lim_{\delta x \to 0}\frac{u(x+\delta x)-u(x)}{\delta x} You might think we are stuck here, but we are able to express the change in u in terms of x: \lim_{\delta x \to 0}\frac{y(u(x)+u(x+\delta x)-u(x))-y(u)}{u(x+\delta x)-u(x)} \cdot \frac{u(x+\delta x)-u(x)}{\delta x} I hope you can see the pattern: \lim_{\delta x \to 0}\frac{y(u(x)+u(x+\delta x)-u(x))-y(u)}{\delta x}=\lim_{\delta x\to 0}\frac{y(u(x+\delta x))-y(u(x))}{\delta x} But this is simply the definition of the derivative of y(u(x)), proving the chain rule." Here's the fallacious step - On the lines "I hope you can see the pattern", you cancel out u(x+\delta x)-u(x) from the top and bottom. But you don't know what function u is - and it's perfectly possible that u(x+\delta x)=u(x) for some or possibly all \delta x . 0 divided by 0 is not 1. I've removed it from the page.--Acer4666 (talk) 18:47, June 2, 2012 (UTC) :If you are still confused, perhaps this will help: you talked above about "treating "du" as an infinitesimal ''number rather than ''zero". But what you need to remember is that u is a function of x, so actually du is the resulting change in u incurred by an infinitesimal change in x. Take u to be the constant function and you will see that it's perfectly possible that du is zero, despite it representing an infinitesimal change (in x)--Acer4666 (talk) 18:57, June 2, 2012 (UTC)